1.3 If \( \left(A^{T}+1\right)^{-1}=\left[\begin{array}{cc}-1 & 2 \\ -3 & 7\end{array}\right] \) is singular if (a) The matrix is not invertable b) \( \left[\begin{array}{cc}-6 & 2 \\ -3 & 2\end{array}\right] \) c) \( \left[\begin{array}{cc}8 & -2 \\ 3 & 0\end{array}\right] \) d) \( \left[\begin{array}{cc}-6 & -3 \\ 2 & 2\end{array}\right] \) e) None of the above

To determine when the matrix \( \left(A^{T}+1\right)^{-1} \) is singular, we need to assess if its determinant is zero. A matrix is singular if it is not invertible, meaning its inverse does not exist. Now, let's analyze the offered matrices. The first matrix can be rewritten in terms of its determinant, which will help us identify singular cases. For options (b), (c), and (d), we can calculate the determinants easily: (b) For \( \left[\begin{array}{cc}-6 & 2 \\ -3 & 2\end{array}\right] \), the determinant is calculated as \((-6)(2) - (2)(-3) = -12 + 6 = -6\) (not singular!). (c) For \( \left[\begin{array}{cc}8 & -2 \\ 3 & 0\end{array}\right] \), the determinant calculates to \(8 \cdot 0 - (-2) \cdot 3 = 0 + 6 = 6\) (not singular). (d) For \( \left[\begin{array}{cc}-6 & -3 \\ 2 & 2\end{array}\right] \), the determinant is calculated as \((-6)(2) - (-3)(2)= -12 + 6 = -6\) (not singular). Thus, the correct answer is (e) None of the above, since all matrices listed do not make \( \left(A^{T}+1\right)^{-1} \) singular.